The length of a given cylindrical wire is increased by 100 %. Due to the consequent decrease in diameter the change in the resistance of

Q: The length of a given cylindrical wire is increased by $100 \%$. Due to the consequent decrease in diameter the change in the resistance of the wire will be .................. $\%$

a If suppose initial length ${l_1} = 100$ then ${l_2} = 100 + 100 = 200$ $\frac{{{R_1}}}{{{R_2}}} = {\left( {\frac{{{l_1}}}{{{l_2}}}} \right)^2} = {\left( {\frac{{100}}{{200}}} \right)^2}$ $ \Rightarrow $ ${R_2} = 4{R_1}$ $\frac{{\Delta R}}{R} \times 100 = \frac{{{R_2} - {R_1}}}{{{R_1}}} \times 100 = \frac{{4{R_1} - {R_1}}}{{{R_1}}} \times 100$$ = 300\% .$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

The length of a given cylindrical wire is increased by $100 \%$. Due to the consequent decrease in diameter the change in the resistance of the wire will be .................. $\%$

A$300$
B$200$
C$100$
D$50$

AIEEE 2003, Medium

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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